CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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$∣ B_{n} ∣ = C_{n}$

Statement

For every $n \in N$ the set $B_{n}$ of balanced bracket words with $n$ pairs of brackets (Balanced bracket words, defined by the recursive grammar) is finite with

$∣ B_{n} ∣ = C_{n},$

the Catalan number (The Catalan number $C_{n} : = ∣ D_{n} ∣$ ).

Facts & Assumptions

Given: a natural number $n$ .

[F1]

The alphabet bijection $(\mapsto U$ , $) \mapsto D$ carries $B_{n}$ onto the set of ballot words of length $2 n$ ( $B_{n}$ is exactly the set of words of length $2 n$ over ${(,)}$ in which every prefix has at least as many $($ as $)$ and the totals are equal).

[F2]

$B_{n}$ is finite (Balanced bracket words, defined by the recursive grammar).

[L1]

$D_{n}$ corresponds bijectively, through step words, to the set of ballot words of length $2 n$ (Dyck paths of semilength $n$ ).

[L2]

$C_{n} = ∣ D_{n} ∣$ (The Catalan number $C_{n} : = ∣ D_{n} ∣$ ).

[L3]

If $A$ is finite and $f : A \to B$ is a bijection then $B$ is finite and $∣ B ∣ = ∣ A ∣$ (The cardinality $∣ A ∣$ of a finite set).

Proof

technique · direct

1.1

By [F1] and [L1] the composite of the alphabet bijection with the inverse of the step-word bijection is a bijection $B_{n} \to D_{n}$ , a composite of two bijections being one.

F1L1

2.1

By [F2] the set $B_{n}$ is finite, so [L3] transports its cardinality along the bijection of step 1.1 and gives $∣ B_{n} ∣ = ∣ D_{n} ∣$ , which is $C_{n}$ by [L2]. At $n = 0$ both sides are $1$ and at $n = 1$ both are $1$ .

F2L2L3step 1.1∎

Remarks

The content is in the theorem above, not here. Once the grammar and the prefix condition are known to describe the same words, the count is a transport along a bijection of alphabets. What makes the corollary worth stating is that it is the first of the three Catalan families whose members are not paths.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

D. Guichard, An Introduction to Combinatorics and Graph Theory, §3.5 (standard reference, not scraped)

∣Bn∣=Cn∣Bn​∣=Cn​